Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{3z + 21}{z - 9} \times \dfrac{5z^2 - 60z + 135}{z^2 + 13z + 42} $
Solution: First factor out any common factors. $q = \dfrac{3(z + 7)}{z - 9} \times \dfrac{5(z^2 - 12z + 27)}{z^2 + 13z + 42} $ Then factor the quadratic expressions. $q = \dfrac {3(z + 7)} {z - 9} \times \dfrac {5(z - 9)(z - 3)} {(z + 7)(z + 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {3(z + 7) \times 5(z - 9)(z - 3) } {(z - 9) \times (z + 7)(z + 6) } $ $q = \dfrac {15(z - 9)(z - 3)(z + 7)} {(z + 7)(z + 6)(z - 9)} $ Notice that $(z + 7)$ and $(z - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {15(z - 9)(z - 3)\cancel{(z + 7)}} {\cancel{(z + 7)}(z + 6)(z - 9)} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $q = \dfrac {15\cancel{(z - 9)}(z - 3)\cancel{(z + 7)}} {\cancel{(z + 7)}(z + 6)\cancel{(z - 9)}} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $q = \dfrac {15(z - 3)} {z + 6} $ $ q = \dfrac{15(z - 3)}{z + 6}; z \neq -7; z \neq 9 $